Remove elements by index in haskell

Question

I\'m new in haskell and I\'m looking for some standard functions to work with lists by indexes.

My exact problem is that i want to remove 3 elements after every 5. If i

Answer 1

the take and drop functions may be able to help you here.

drop, take :: Int -> [a] -> [a]

from these we could construct a function to do one step.

takeNdropM :: Int -> Int -> [a] -> ([a], [a])
takeNdropM n m list = (take n list, drop (n+m) list)

and then we can use this to reduce our problem

takeEveryNafterEveryM :: Int -> Int -> [a] -> [a]
takeEveryNafterEveryM n m [] = []
takeEveryNafterEveryM n m list = taken ++ takeEveryNafterEveryM n m rest
    where
        (taken, rest) = takeNdropM n m list

*Main> takeEveryNafterEveryM 5 3 [1..20]
[1,2,3,4,5,9,10,11,12,13,17,18,19,20]

---

since this is not a primitive form of recursion, it is harder to express this as a simple fold.

so a new folding function could be defined to fit your needs

splitReduce :: ([a] -> ([a], [a])) -> [a] -> [a]
splitReduce f []   = []
splitReduce f list = left ++ splitReduce f right
    where
        (left, right) = f list

then the definition of takeEveryNafterEveryM is simply

takeEveryNafterEveryM2 n m = splitReduce (takeNdropM 5 3)